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\title{\heiti\zihao{2} 习题6.3}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{引理1}
一个分式 $\frac{f}{g}$ 的次数如果小于 $0,$ 并且它是既约分式,那么称它为真分式.

存在性:

$K(x)$ 中每一个分式都可以唯一地表示成一个多项式与一个真分式的和. 证明 把分式 $\frac{f}{g}$ 的分子与分母消去它们的最大公因式后,就变成既约分式.因此不 妨设 $\frac{f}{g}$ 是一个既约分式,对 $f$ 与 $g$ 作带余除法得
\begin{equation}
    f=g h+r, \quad \operatorname{deg} r<\operatorname{deg} g 
\end{equation}

于是
\begin{equation}
\frac{f}{g}=\frac{g h+r}{g}=\frac{g h}{g}+\frac{r}{g}=\frac{h}{1}+\frac{r}{g}=h+\frac{r}{g}\qquad(2)
\end{equation}

由于 $(f, g)=1,$ 因此从 (42) 式得 $,(g, r)=(f, g)=1$ .于是 $\frac{r}{g}$ 是一个既约分式,又由于
$\operatorname{deg}r-\operatorname{deg} g<0,$ 因此 $\frac{r}{g}$ 是一个真分式.存在性证毕.

唯一性:

假设还有 $\frac{f}{g}=h_{1}+\frac{r_{1}}{g_{1}},$ 其中 $h_{1}, r_{1}, g_{1} \in K[x], \operatorname{deg} r_{1}<\operatorname{deg} g_{1},$且$a \left(r_{1}, g_{1}\right)=1,$ 则
$$
h-h_{1}=\frac{r_{1}}{g_{1}}-\frac{r}{g}=\frac{r_{1} g-r g_{1}}{g_{1} g}a
$$

假如 $h \neq h_{1},$ 则 $\operatorname{deg}\left(h-h_{1}\right) \geqslant 0 .$ 然而
$$
\operatorname{deg} \frac{r_{1} g-r g_{1}}{g_{1} g}=\operatorname{deg}\left(r_{1} g-r g_{1}\right)-\operatorname{deg}\left(g_{1} g\right)<0
$$

矛盾.因此 $h=h_{1},$ 从而 $r_{1} g-r g_{1}=0 .$ 由此得出 $, \frac{r}{g}=\frac{r_{1}}{g_{1}}$ .唯一性证毕.

\section{引理2——有理真分式的因式分解定理}
若有$f(x),g(x) \in K[x]$且$\deg g(x)>0$(最高次次数),设$g(x)=p_1^{l_1}(x)p_2^{l_2}(x)\cdots p_m^{l_m}(x)$,其中$p_i^{l_i}(x)$互相不可约,则若有$\deg f(x)< \deg g(x)$,则存在$A_{ij_i}(x) \in K[x]$,且$\deg A_{ij_i}(x)<\deg p_i(x),i=1,2,\cdots ,m,j_i=1,2,\cdots,l_i$,满足:
$$
\frac{f(x)}{g(x)}=\sum_{i=1}^{m} \sum_{j_{i}=1}^{l_i} \frac{A_{i j{i}}(x)}{p^{j_i}_{i}(x)}
$$

证明:
$B_{i}(x)=p_{1}^{l_1}(x) \cdots p_{i-1}^{l_{i-1}}(x) p_{i+1}^{l_{i+1}}(x) \cdots p_{m}^{l_m}(x)$

其中 $i=1,2, \cdots, m$ 则 $\left(B_{1}(x), B_{2}(x), \cdots, B_{m}(x)\right)=1 .$ 从而存在 $u_{i}(x) \in K[x], i=1,2 ,\cdots, m$ 使得

\begin{equation}
u_1(x)B_1(x)+u_2(x)B_2(x)+\cdots +u_m(x)B_m(x)=f(x)
\end{equation}
(可由裴蜀定理易知)

从而对$u_i(x)$和$p_i^{l_i}(x)$做带余除法,得:

$u_i(x)=h_i(x)p_i^{l_i}(x)+r_i(x)$,其中$\deg r_i(x)< \deg p_i^{l_i}(x),i=1,2,\cdots ,m$

将其带入(3)式,得
\begin{equation}
    \sum_{i=1}^{m}\left[h_{i}(x) p_{i}^{l_i}(x) B_{i}(x)+r_{i}(x) B_{i}(x)\right]=f(x)
\end{equation}

即
\begin{equation}
\left[\sum_{i=1}^{m}h_{i}(x)\right] g(x) +\sum_{i=1}^{m}r_{i}(x) B_{i}(x)=f(x) \qquad(3)
\end{equation}

(5)式两边同时除以$g(x)$,得
$$
\frac{f(x)}{g(x)}=\sum_{i=1}^{m}h_{i}(x)+\sum_{i=1}^{m}\frac{r_i(x)}{p_i^{l_i}(x)}
$$

由于$\deg f(x)< \deg g(x)$,且不妨设$(f(x),g(x))=1$,所以$\frac{f(x)}{g(x)}$是真分式,由引理1唯一性知,$\sum_{i=1}^{m}h_{i}(x)=0$

从而
$$
\frac{f(x)}{g(x)}=\sum_{i=1}^{m}\frac{r_i(x)}{p_i^{l_i}(x)}
$$

由于$\deg r_i (x)<\deg p_i^{l_i}(x)$,所以存在$A_{ij_i}(x) \in K[x]$,

且$\deg A_{ij_i}(x)<\deg p_i(x),j_i=1,2,\cdots,l_i$,使得
$$
\frac{r_i(x)}{p_i^{l_i}(x)}=\sum_{j_i=1}^{l_i}\frac{A_{ij_i}(x)}{p_i^{j_i}(x)},i=1,2,\cdots,m
$$

从而有
$$
\frac{f(x)}{g(x)}=\sum_{i=1}^{m} \sum_{j_{i}=1}^{l_i} \frac{A_{i j{i}}(x)}{p^{j_i}_{i}(x)}
$$

证毕.

\section{计算下列不定积分}
\subsection{$\int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} \mathrm{~d} x(a>0)$}
\textbf{解}\quad
令$x=a\sin u,u=\arcsin \frac{x}{a}$

$$
\begin{aligned}
    \int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} \mathrm{~d} x&=\int \frac{1}{a^{2} \sin ^{2} u \cos u} \cdot a \cos u \mathrm{~d} u\\&=\frac{1}{a^{2}} \int \frac{1}{\sin ^{2} u} \mathrm{~d} u\\&=\frac{1}{a^{2}}(-\cot u)+C\\&=-\frac{\sqrt{a^{2}-x^{2}}}{a^{2} x}+C
\end{aligned}
$$


\subsection{$\int \frac{1}{x^{2} \sqrt{b^{2}+x^{2}}} \mathrm{~d} x(b>0)$}
\textbf{解}\quad
令$x=b\tan u$

$$
\begin{aligned}
    \int \frac{1}{x^{2} \sqrt{x^{2}+b^{2}}} \mathrm{~d} x&=\int \frac{1}{b^{2} \tan ^{2} u \cdot b \sec u}b\sec^2 u \mathrm{~d}u \\&=\int \frac{\sec u \mathrm{~d} u}{b^{2} \tan ^{2} u} \mathrm{~d} u=\frac{1}{b^{2}} \int \frac{\cos u}{\sin ^{2} u} \mathrm{~d} u\\&=\frac{1}{b^{2}} \int \frac{\mathrm{~d} \sin u}{\sin ^{2} u}=-\frac{1}{b^{2}}\left(\frac{1}{\sin u}\right)+C\\&=-\frac{\sqrt{b^{2}+x^{2}}}{b^{2} x}+C
\end{aligned}
$$


\subsection{$\int \frac{x^{5}}{\sqrt{1-x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{x^{5}}{\sqrt{1-x^{2}}} \mathrm{~d} x&=\frac{1}{2} \int \frac{x^{4} \mathrm{~d} x^{2}}{\sqrt{1-x^{2}}}\\&=\frac{1}{2} \int \frac{u^{2} \mathrm{~d} u}{\sqrt{1-u}}\\&=\frac{1}{2} \int \frac{\left(1-v^{2}\right)^{2}}{v} \cdot(-2 v) \mathrm{~d} v=\int-1+2 v^{2}-v^{4} \mathrm{~d} v\\&=-v+\frac{2}{3} v^{3}-\frac{1}{5} v^{5}+C\\&=-\sqrt{1-x^{2}}+\frac{2}{3} \sqrt{\left(1-x^{2}\right)^{3}}-\frac{1}{5} \sqrt{\left(1-x^{2}\right)^{5}}+C
\end{aligned}
$$

\subsection{$\int \frac{\sqrt{x}}{1-\sqrt[3]{x}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{\sqrt{x}}{1-\sqrt[3]{x}} \mathrm{~d} x&=\int \frac{u^{3}}{1-u^{2}} \mathrm{~d} u^{6}\\&=6 \int \frac{u^{8}}{1-u^{2}} \mathrm{~d} u=6 \int-u^{6}-u^{4}-u^{2}-1+\frac{1}{1-u^{2}} \mathrm{~d} u\\&=6\left(-\frac{u^{7}}{7}-\frac{u^{5}}{5}-\frac{n^{3}}{3}-u+\ln \sqrt{|\mathrm{~d} \frac{1+u}{1-u}} |\mathrm{~d}\right)+C\\&=-\frac{6}{7} x^{\frac{7}{6}}-\frac{6}{5} x^{\frac{5}{6}}-2 \sqrt{x}-6 x^{\frac{1}{6}}+6 \ln \sqrt{\frac{1+x^{\frac{1}{6}}}{1-x^{\frac{1}{6}}}}+C
\end{aligned}
$$

\subsection{$\int \frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{\mathrm{e}^{2 x}+\mathrm{e}^{-2 x}+1} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{e^{x}-e^{-x}}{e^{2 x}+e^{-2 x}+1} \mathrm{~d} x&=\int \frac{e^{2 x}-1}{e^{4 x}+e^{2 x}+1} \mathrm{~d} e^{x}=\int \frac{u^{2}-1}{u^{4}+u^{2}+1} \mathrm{~d} u\\&=\frac{1}{2} \int \frac{2 u+1}{u^{2}+u+1}-\frac{2 u-1}{u^{2}-u+1} \mathrm{~d} u\\&=\frac{1}{2}\left(\int \frac{\mathrm{~d} u^{2}+u+1}{u^{2}+u+1}-\frac{\mathrm{~d} u^{2}-u+1}{u^{2}-u+1}\right)\\&=\frac{1}{2}\left(\ln \left|u^{2}+u+1\right|-\ln \left|u^{2}-u+1\right|\right)\\&=\frac{1}{2} \ln \left|\frac{e^{2 x}+e^{x}+1}{e^{2+}-e^{x}+1}\right|
\end{aligned}
$$


\subsection{$\int \frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+\mathrm{e}^{-x}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{e^{x}}{e^{x}+e^{-x}} \mathrm{~d} x=\int \frac{u}{u+\frac{1}{u}} \frac{\mathrm{~d} u}{u}=\frac{1}{2} \int \frac{\mathrm{~d} u^{2}+1}{u^{2}+1}=\frac{1}{2} \ln \left|u^{2}+1\right|+C=\frac{1}{2} \ln \left(e^{2 x}+1\right)+C$$


\subsection{$\int \frac{1}{1+\tan x} \mathrm{~d} x $}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{1}{1+\tan x} \mathrm{~d} x&=\int \frac{1}{1+u} \cdot \frac{1}{1+u^{2}} \mathrm{~d} u\\&=\frac{1}{2} \int \frac{1}{1+u}+\frac{1-u}{1+u^{2}} \mathrm{~d} u\\&=\frac{1}{2}\left(\ln |1+u|+\arctan u-\frac{1}{2} \ln \left|u^{2}+1\right|\right)\\&=\frac{1}{2} \ln |1+\tan x|+\frac{x}{2}+\frac{1}{2} \ln |\cos x|+C
\end{aligned}
$$

\subsection{$\int \frac{1}{x^{3}-2 x+1} \mathrm{~d} x $}
\textbf{解}\quad
$$\int \frac{1}{x^{3}-2 x+1} \mathrm{~d} x=\int \frac{1}{x-1}+\frac{-x-2}{x^{2}+x-1} \mathrm{~d} x 
=\int \frac{1}{x-1} \mathrm{~d} x \quad-\int \frac{x+2}{x^{2}+x-1} \mathrm{~d} x$$

$$\int \frac{1}{x-1} \mathrm{~d}x=\ln |x-1 |+C$$

$$
\begin{aligned}
\int \frac{x}{x^{2}+x-1} \mathrm{~d} x&=\int \frac{x}{\left(x+\frac{1}{2}\right)^{2}-\frac{5}{4}} \mathrm{~d} x\\&=\int \frac{4 x}{4\left(x+\frac{1}{2}\right)^{2}-5} \mathrm{~d} x 
\\&=\int \frac{4 u-2}{4 u^{2}-5} \mathrm{~d} u\\&=2 \int \frac{2 u}{4 u^{2}-5}-\frac{1}{4u^{2}-5} \mathrm{~d} u
\end{aligned}
$$

$$\int \frac{2u}{4 n^{2}-5} \mathrm{~d} u=\frac{1}{4} \ln \left|4 u^{2}-5\right|$$

$$\int \frac{1}{4 u^{2}-5} \mathrm{~d} u=\frac{1}{4 \sqrt{5}}\left(\ln \left|\frac{2 u-\sqrt{5}}{2 u+\sqrt{5}}\right|\right)$$

$$\int \frac{4 u-2}{4 u^{2}-5} \mathrm{~d} u=\frac{1}{2} \ln \left|4 u^{2}-5\right|-\frac{1}{2 \sqrt{5}}\left(\ln \left|\frac{2 u-\sqrt{5}}{2 u+\sqrt{5}}\right|\right)$$

$$\int \frac{1}{x^{3}-2 x+1} \mathrm{~d} x=\frac{1}{2} \ln \left|x^{2}+x-1\right|-\frac{1}{2 \sqrt{5}}\left(\ln \left|\frac{2 x+1-\sqrt{5}}{2 x+1+\sqrt{5}}\right|\right)+C$$


\subsection{$\int \frac{1}{1+x^{4}} \mathrm{~d} x$}
\textbf{解}$1^{\circ}$\quad
$$\int \frac{1}{x^4+1}\mathrm{~d}x=\int \frac{1}{2}\left(\frac{x^{2}+1}{x^{4}+1}-\frac{x^{2}-1}{x^{4}+1}\right)\mathrm{~d}x $$

$$\int \frac{x^{2}+1}{x^{4}+1} \mathrm{~d} x=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} \mathrm{~d} x=\int \frac{\mathrm{\mathrm{~d}}\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+2}=a+C$$

$$\int \frac{x^{2}-1}{x^{4}+1} \mathrm{~d} x=\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}} \mathrm{~d} x=\int \frac{\mathrm{\mathrm{~d}}\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-2}=\frac{1}{2 \sqrt{2}} \ln \left(\frac{x^{2}-x \sqrt{2}+1}{x^{2}+x \sqrt{2}+1}\right)+C$$

$$\int \frac{1}{x^4+1}\mathrm{~d}x=\int \frac{1}{2}\left(\frac{x^{2}+1}{x^{4}+1}-\frac{x^{2}-1}{x^{4}+1}\right)\mathrm{~d}x =\frac{\sqrt{2}}{4} \arctan \frac{x^{2}-1}{x \sqrt{2}}+\frac{1}{4 \sqrt{2}} \ln \left(\frac{x^{2}+x \sqrt{2}+1}{x^{2}-x \sqrt{2}+1}\right)+C$$

\textbf{解}$2^{\circ}$\quad
$$
\begin{aligned}
    \int \frac{\mathrm{\mathrm{~d}} x}{x^{4}+1}&=\int \frac{\frac{\sqrt{2}}{4} x+\frac{1}{2}}{x^{2}+x \sqrt{2}+1} \mathrm{~d} x+\int \frac{-\frac{\sqrt{2}}{4} x+\frac{1}{2}}{x^{2}-x \sqrt{2}+1} \mathrm{~d} x\\&=\frac{\sqrt{2}}{4} \int \frac{\left(x+\frac{\sqrt{2}}{2}\right) \mathrm{\mathrm{~d}} x}{\left(x+\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}+\frac{1}{4} \int \frac{\mathrm{\mathrm{~d}} x}{\left(x+\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}\\&-\frac{\sqrt{2}}{4} \int \frac{\left(x-\frac{\sqrt{2}}{2}\right) \mathrm{\mathrm{~d}} x}{\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}+\frac{1}{4} \int \frac{\mathrm{\mathrm{~d}} x}{\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{1}{2}}\\&=\frac{1}{4 \sqrt{2}} \ln \frac{x^{2}+x \sqrt{2}+1}{x^{2}-x \sqrt{2}+1}+\frac{\sqrt{2}}{4} \arctan \left(\frac{x \sqrt{2}}{1-x^{2}}\right)+C
\end{aligned}
$$


\subsection{$\int \frac{\mathrm{\mathrm{~d}} x}{x^{4}\left(1+x^{2}\right)} $}
\textbf{解}\quad
$$\begin{aligned} \int \frac{\mathrm{~d} x}{x^{4}\left(1+x^{2}\right)} &=\int\left[\frac{1}{x^{4}}-\frac{1}{x^{2}\left(1+x^{2}\right)}\right] \mathrm{\mathrm{~d}} x \\ &=\int\left[\frac{1}{x^{4}}-\frac{1}{x^{2}}+\frac{1}{\left(1+x^{2}\right)}\right] \mathrm{\mathrm{~d}} x=-\frac{1}{3 x^{3}}+\frac{1}{x}+\arctan x+C \end{aligned}$$


\subsection{$\int \frac{1}{x^{11}+2 x} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{1}{x^{11}+2 x}&=\int \frac{1}{x\left(x^{10}+2\right)} \mathrm{~d} x\\&=\int \frac{x^{9}}{x^{10}\left(x^{10}+2\right)} \mathrm{~d} x=\frac{1}{10} \int \frac{\mathrm{~d} x^{10}}{x^{10}\left(x^{10}+2\right)}\\&=\frac{1}{10} \int \frac{\mathrm{~d} u}{u(u+2)}+C=\frac{1}{20} \ln \left|\frac{u}{u+2}\right|\\&=\frac{1}{20} \ln \left|\frac{x^{10}}{x^{10}+2}\right|+C
\end{aligned}
$$


\subsection{$\int \frac{x^{4}+1}{x^{6}+1} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int \frac{x^{4}+1}{x^{6}+1}=\int \frac{1}{x^{2}+1} \mathrm{~d} x+\int \frac{x^{2}}{x^{6}+1} \mathrm{~d} x=\int \frac{1}{x^{2}+1} \mathrm{~d} x+\frac{1}{3} \int \frac{\mathrm{~d} x^{3}}{x^{6}+1}=\arctan x+\frac{1}{3} \arctan x^{3}+C
$$

\section{计算下列不定积分}
\subsection{$\int \frac{\mathrm{\mathrm{~d}} x}{1+\sin x+\cos x}$}
\textbf{解}\quad
$\int \frac{1}{1+\sin x+\cos x} \mathrm{~d} x=\int\frac{1}{1+\frac{2 u}{1+u^{2}}+\frac{1-u^{2}}{1+n^{2}}}\frac{2}{1+u^{2}} \mathrm{~d} u=\int \frac{\mathrm{~d} u+1}{u+1}=\ln \left|\tan \frac{x}{2}+1\right|+C$


\subsection{$\int \frac{\mathrm{\mathrm{~d}} x}{2+\sin x}$}
\textbf{解}\quad
$$
\begin{aligned}
\int \frac{1}{2+\sin x} \mathrm{~d} x&=\int \frac{1}{u^{2}+u+1} \mathrm{~d} u=\int \frac{1}{\left(u+\frac{1}{2}\right)^{2}+\frac{3}{4}} \mathrm{~d} u=\int \frac{1}{v^{2}+\frac{3}{4}} \mathrm{~d} v\\&=4 \int \frac{1}{4 v^{2}+3} \mathrm{~d} v=\frac{2}{\sqrt{3}} \arctan \frac{2}{\sqrt{3}} v\\&=\frac{2}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\left(2 \tan \frac{x}{2}+1\right)\right)+C
\end{aligned}
$$

\subsection{$\int \frac{x+2}{x^{2} \sqrt{1-x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{x+2}{x^{2} \sqrt{1-x^{2}}} \mathrm{~d} x=\int \frac{1}{x \sqrt{1-x^{2}}}+\frac{2}{x^{2} \sqrt{1-x^{2}}} \mathrm{~d} x$$
$$
\begin{aligned}
\int \frac{1}{x \sqrt{1-x^{2}}} \mathrm{~d} x=\int \frac{1}{u^{2}-1} \mathrm{~d} u=\frac{1}{2} \int \frac{1}{u-1}-\frac{1}{u+1} \mathrm{~d} u=\frac{1}{2} \ln \left|\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}\right|
\end{aligned}
$$
$$\int \frac{1}{x^{2} \sqrt{1-x^{2}}} \mathrm{~d} x=\int \frac{1}{\sin ^{2} u} \mathrm{~d} u=-\cot u=-\frac{\sqrt{1-x^{2}}}{x}$$
$$\int \frac{x+2}{x^{2} \sqrt{1-x^{2}}}=\frac{1}{2} \ln \left|\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}+1}}\right|-\frac{2 \sqrt{1-x^{2}}}{x}+C$$

\subsection{$\int \frac{x+1}{x^{2} \sqrt{x^{2}-1}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
\int \frac{x+1}{x^{2} \sqrt{x^{2}-1}} \mathrm{~d} x&=\int \frac{\mathrm{~d} x}{x \sqrt{x^{2}-1}}+\int \frac{\mathrm{~d} x}{x^{2} \sqrt{x^{2}-1}}\\
\int \frac{\mathrm{~d} x}{x \sqrt{x^{2}-1}}&=\int \frac{\sqrt{x^{2}-1}}{x} \cdot \frac{1}{x \sqrt{x^{2}-1}} \mathrm{~d} \sqrt{x^{2}-1}\\&=\int \frac{1}{u^{2}+1} \mathrm{~d} u\\&=\arctan u+C\\&=\arctan \sqrt{x^{2}-1}+C\\
\int \frac{\mathrm{~d} x}{x^{2} \sqrt{x^{2}-1}}&=\int \frac{\sqrt{x^{2}-1}}{x} \frac{1}{x^{2} \sqrt{x^{2}-1}} \mathrm{~d} \sqrt{x^{2}-1}\\&=\int \frac{1}{\sqrt{u^{2}+1}\left(u^{2}+1\right)} \mathrm{~d} u
\end{aligned}
$$

令$u=\tan v$

$$\int \frac{1}{\left(1+u^{2}\right)^{\frac{3}{2}}} \mathrm{~d} u=\int \cos v \mathrm{~d} v=\sin v+C=\frac{\sqrt{x^{2}-1}}{x}$$

$$\int \frac{x+1}{x^{2} \sqrt{x^{2}-1}} \mathrm{~d} x=\arctan \sqrt{x^2-1}+\frac{\sqrt{x^2-1}}{x}+C$$


\subsection{$\int \frac{1-\sqrt{x+2}}{1+\sqrt[3]{x+2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\begin{aligned}
    \int \frac{1-\sqrt{x+1}}{1+\sqrt[3]{x+1}} \mathrm{~d} x&=6 \int \frac{u^{5}\left(1-u^{3}\right)}{1+u^{2}} \mathrm{~d} u\\&=6 \int\left(-u^{6}+u^{4}+u^{3}-u^{2}-u+1+\frac{u-1}{1+u^{2}}\right) \mathrm{\mathrm{~d}} u\\&=-\frac{6}{7}(x+2)^{\frac{7}{6}}+\frac{6}{5}(x+2)^{\frac{5}{6}}+\frac{3}{2}(x+2)^{\frac{2}{3}}-2 \sqrt{x+2}\\&-3(x+2)^{\frac{1}{3}}+6(x+2)^{\frac{1}{6}}+3 \ln \left(1+(x+2)^{\frac{1}{3}}\right)-6 \arctan (x+2)^{\frac{1}{6}}
\end{aligned}
$$


\subsection{$\int \frac{1}{\sqrt[3]{(x-2)^{2}(x+1)}} \mathrm{\mathrm{~d}} x$}
\textbf{解}\quad
令$\sqrt[3]{\frac{x+1}{x-2}}=u, x=\frac{2u^{3}+1}{u^{3}-1}$,$\mathrm{~d} x=-\frac{9 u^{2}}{\left(u^{3}-1\right)^{2}}\mathrm{~d}u$

$$
\begin{aligned}
\int \frac{1}{\sqrt[3]{(x-2)^{2}(x+1)}} \mathrm{~d} x&=-\int \frac{1}{x-2} \cdot \frac{1}{u} \cdot \frac{9 u^{2}}{\left(u^{3}-1\right)^{2}} \mathrm{~d} u\\&=-\int \frac{1}{x-2} \cdot \frac{1}{u} \cdot \frac{9u ^{2}(x-2)^2}{9} \mathrm{~d} u=-\int u(x-2) \mathrm{~d} u\\&= 3 \int \frac{u}{1-u^{3}} \mathrm{~d} u\\\\
\int \frac{u}{1-u^{3}} \mathrm{~d} u&=-\int \frac{1}{3(u-1)}+\frac{1-u}{3\left(u^{2}+u+1\right)} \mathrm{~d} u\\
\int \frac{1}{3(u-1)} \mathrm{~d} u&=\frac{1}{3} \ln |u-1|\\
\int \frac{1-u}{3\left(u^{2}+u+1\right)}&=\frac{1}{3} \int \frac{1-u}{\left(u+\frac{1}{2}\right)^{2}+\frac{3}{4}} \mathrm{~d} u\\&=\frac{2}{3} \int \frac{3-2v}{4 v^{2}+3} \mathrm{~d} v=\frac{2}{3} \int \frac{\sqrt{3}-w}{2\left(w^{2}+1\right)} \mathrm{~d} w\\&=\frac{1}{3}\left(\int \frac{-w}{w^{2}+1} \mathrm{~d} w+\int \frac{\sqrt{3}}{w^{2}+1} \mathrm{~d} w\right)\\&=-\frac{1}{3}\left(\frac{1}{2} \ln \left|w^{2}+1\right|-\sqrt{3} \arctan w\right)\\&=-\frac{1}{6} \ln \left|u^2+u+1\right|+\frac{\sqrt{3}}{3} \arctan \frac{2 u+1}{\sqrt{3}}+C\\
3\int \frac{u}{1-u^{3}} \mathrm{~d} u&=- \ln |u-1|+\frac{1}{2} \ln \left|u^2+u+1\right|-{\sqrt{3}} \arctan \frac{2 u+1}{\sqrt{3}}+C
\end{aligned}
$$


$$
\begin{aligned}
\therefore\int \frac{1}{\sqrt[3]{(x-2)^{2}(x+1)}} \mathrm{~d} x&=-\ln \left|\sqrt[3]{\frac{x+1}{x-2}}-1\right|+\frac{1}{2} \ln \left(\left|1+\sqrt[3]{\frac{x+1}{x-2}}+\left(\sqrt[3]{\frac{x+1}{x-2}}\right)^{2}\right|\right)\\&-\sqrt{3} \arctan \left( \frac{2 \sqrt[3]{\frac{x+1}{x-2}}+1}{\sqrt{3}} \right)+C
\end{aligned}
$$

\end{document}